If ABC is an isosceles triangle and the coord | vedclass

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(C) Let the coordinates of vertex $A$ be $(x, y)$.Since $	riangle ABC$ is isosceles with base $BC$, we have $AB = AC$, which implies $AB^2 = AC^2$.Us

Vedclass · Accepted Answer

$(\frac{5}{6}, 6)$

Vedclass · Answer

(C) Let the coordinates of vertex $A$ be $(x, y)$.Since $	riangle ABC$ is isosceles with base $BC$, we have $AB = AC$, which implies $AB^2 = AC^2$.Using the distance formula: $(x - 1)^2 + (y - 3)^2 = (x + 2)^2 + (y - 7)^2$.Expanding both sides: $x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 14y + 49$.Simplifying: $-2x - 6y + 10 = 4x - 14y + 53$.Rearranging terms: $8y - 6x = 43$.We check the given options to see which point satisfies the equation $8y - 6x = 43$.For option $C$: $8(\frac{5}{6}) - 6(6) = \frac{20}{3} - 36 = \frac{20 - 108}{3} = -\frac{88}{3} 
eq 43$.For option $B$: $8(5) - 6(-\frac{1}{8}) = 40 + \frac{3}{4} = 40.75 
eq 43$.Re-evaluating the options, let's check if there is a typo in the provided options. If we test $A(x, y) = (\frac{5}{6}, 6)$, it does not satisfy the equation. However, checking the equation $8y - 6x = 43$ for the provided options, none perfectly satisfy it. Given the standard nature of such problems, we assume the intended answer is based on the locus $8y - 6x = 43$.

If $ABC$ is an isosceles triangle and the coordinates of the base points are $B(1, 3)$ and $C(-2, 7)$, then the coordinates of $A$ can be:

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